@robjohn, I know how to prove the expression is not equal to a fixed candidate function, such as $\frac{1}{2} \left( x + \sqrt{1 + x^2}\right),$ but I do not know how to answer Why Not at a precalculus level.
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Will JagyOct 20 '12 at 19:19

2 Answers
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Manipulations of this and similar quantities like $ \frac{e^x}{1+e^x}$, or $e^{-x} (1 + e^x)$, and their logarithms, are common in logistic regression in statistics.

There is no closed form simplification but you can expand $\log(1 \pm t) = \pm t + \frac{t^2}{2} \pm \frac{t^2}{3} + \cdots$ in a power series for specially chosen $t$ (smaller than 1 in absolute value) to illustrate how $f(x) = \log (1 + e^x)$ is approximately $x$. Possibilities include

to write $f(x) = x + \log (1 + e^{-x})$ and use $t = e^{-x}$ to get a series valid for positive $x$,